Divide the following complex numbers. $ \dfrac{-23-15i}{-5-i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-5+i}$ $ \dfrac{-23-15i}{-5-i} = \dfrac{-23-15i}{-5-i} \cdot \dfrac{{-5+i}}{{-5+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-23-15i) \cdot (-5+i)} {(-5-i) \cdot (-5+i)} = \dfrac{(-23-15i) \cdot (-5+i)} {(-5)^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-23-15i) \cdot (-5+i)} {(-5)^2 - (-1i)^2} = $ $ \dfrac{(-23-15i) \cdot (-5+i)} {25 + 1} = $ $ \dfrac{(-23-15i) \cdot (-5+i)} {26} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-23-15i}) \cdot ({-5+i})} {26} = $ $ \dfrac{{-23} \cdot {(-5)} + {-15} \cdot {(-5) i} + {-23} \cdot {1 i} + {-15} \cdot {1 i^2}} {26} $ Evaluate each product of two numbers. $ \dfrac{115 + 75i - 23i - 15 i^2} {26} $ Finally, simplify the fraction. $ \dfrac{115 + 75i - 23i + 15} {26} = \dfrac{130 + 52i} {26} = 5+2i $